How does crossing two 3d vectors produce a third one that is perpendicular to both?

Question

Can someone help me understand the cross product a little better, for me it makes more sense for the new vector to be somewhere between the original vectors and closer to the bigger one but that would be just adding to one another right? how does crossing say for example vector $\vec A \times \vec B =\begin{bmatrix} 1 \\ 0 \\ 0 \\\end{bmatrix} \times \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \\\end{bmatrix}= \vec C$?

Answer 1

The cross product can be defined using a determinant $$ \vec{A}\times\vec{B} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat {k} \\ A_1 & A_2 & A_3 \\ B_1 & B_2 & B_3\end{array}\right| $$ So, when dot this result with $\vec{C}$ you get the same as replacing $\hat{i}$ with $C_1$, etc. That means, $$ (\vec{A}\times\vec{B})\cdot\vec{C}=\mbox{det}(\vec{C},\vec{A},\vec{B}), $$ where the notation on the right is the determinant whose first row is $\vec{C}$, whose second row is $\vec{A}$ and whose third row is $\vec{B}$. And you know that determinants of matrices with two identical rows are $0$. That gives you $$ (\vec{A}\times\vec{B})\cdot \vec{A}=0\\ (\vec{A}\times\vec{B})\cdot \vec{B}=0. $$ In other words, the cross product is orthogonal to the original two vectors.